End Correction


by vishmz@gmail.com
Tags: correction
vishmz@gmail.com
#1
Apr19-07, 05:00 AM
P: n/a
In the analysis of sound waves generated in a pipe open at both ends
or open at one end only, the tube length is taken as L=L_0+0.3 D,
where L_0 is the actual tube length and D is the diameter of the tube.
Is this only a thumb rule? Or can a physical argument be given for the
factor 0.3 D? I havent found any argument in undergrad physics books.
Can someone help?

regards.

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Gerard Westendorp
#2
Apr19-07, 05:00 AM
P: n/a
vishmz@gmail.com wrote:

> In the analysis of sound waves generated in a pipe open at both ends
> or open at one end only, the tube length is taken as L=L_0+0.3 D,
> where L_0 is the actual tube length and D is the diameter of the tube.
> Is this only a thumb rule? Or can a physical argument be given for the
> factor 0.3 D? I havent found any argument in undergrad physics books.
> Can someone help?


This is a low frequency approximation, that works quite well if the
wavelength is smaller than the pipe diameter.

At the open end of a pipe, you have velocity fluctuations but little
pressure fluctuations. So the impedance here is mass-like. The mass
impedance of a small slice (dx) of tube is jw rho dx/S. [w is angular
frequency, rho is density, S is cross sectional area, and j = the root
of -1.]

The mass impedance of the open end can be estimated by slicing up space
in concentric spherical slices of radius (r). Each slice has mass
impedance jw rho dr /(4 pi r^2). Integrate this over r from R (radius of
tube) to infinity, and you get jw rho /(4 pi R), the same as a tube
slice of length R/4, or D/8.

So the above estimate gives an end correction of 0.125 D. But it is a
bit crude, since we start integrating from a ball of radius R. In
reality, the pattern is more complex. If you account for the pattern in
more detail, you get a different value. I don't know if it is 0.3, I
thought it was higher.

Note that we used low frequency approximation, so that the whole field
around the tube rnd is behaving like a mass. It is a bit like an open
end has a little ball of air attached to it, whose mass has to be
displaced by the acoustic wave.

An important application of the end effect is the impedance of a small
hole. This is the same as a little pipe of length about 0.8D. You can
use this to design all kinds of resonators and dampers, and understand
how sound can leak through holes.

Gerard

vishmz@gmail.com
#3
Apr20-07, 05:00 AM
P: n/a
On Apr 19, 3:51 am, Gerard Westendorp <west...@xs4all.nl> wrote:
> vis...@gmail.com wrote:
> > In the analysis of sound waves generated in a pipe open at both ends
> > or open at one end only, the tube length is taken as L=L_0+0.3 D,
> > where L_0 is the actual tube length and D is the diameter of the tube.
> > Is this only a thumb rule? Or can a physical argument be given for the
> > factor 0.3 D? I havent found any argument in undergrad physics books.
> > Can someone help?

>
> This is a low frequency approximation, that works quite well if the
> wavelength is smaller than the pipe diameter.
>
> At the open end of a pipe, you have velocity fluctuations but little
> pressure fluctuations. So the impedance here is mass-like. The mass
> impedance of a small slice (dx) of tube is jw rho dx/S. [w is angular
> frequency, rho is density, S is cross sectional area, and j = the root
> of -1.]
>
> The mass impedance of the open end can be estimated by slicing up space
> in concentric spherical slices of radius (r). Each slice has mass
> impedance jw rho dr /(4 pi r^2). Integrate this over r from R (radius of
> tube) to infinity, and you get jw rho /(4 pi R), the same as a tube
> slice of length R/4, or D/8.
>
> So the above estimate gives an end correction of 0.125 D. But it is a
> bit crude, since we start integrating from a ball of radius R. In
> reality, the pattern is more complex. If you account for the pattern in
> more detail, you get a different value. I don't know if it is 0.3, I
> thought it was higher.
>
> Note that we used low frequency approximation, so that the whole field
> around the tube rnd is behaving like a mass. It is a bit like an open
> end has a little ball of air attached to it, whose mass has to be
> displaced by the acoustic wave.
>
> An important application of the end effect is the impedance of a small
> hole. This is the same as a little pipe of length about 0.8D. You can
> use this to design all kinds of resonators and dampers, and understand
> how sound can leak through holes.
>
> Gerard



Dear Gerard,

Thanx for the explanation!!


Hans Werner Strube, strube&physik3*gwdg*de
#4
Apr21-07, 05:00 AM
P: n/a

End Correction


vishmz@gmail.com wrote:
> On Apr 19, 3:51 am, Gerard Westendorp <west...@xs4all.nl> wrote:

...
>> This is a low frequency approximation, that works quite well if the
>> wavelength is smaller than the pipe diameter.

...
> Thanx for the explanation!!


You mean "larger" rather than "smaller" - low frequency corresponds to
large wavelength.

--

Hans Werner Strube strube(@)physik3(.)gwdg(.)de
Drittes Physikalisches Institut, Univ. Goettingen
Friedrich-Hund-Platz 1, 37077 Goettingen, Germany



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